text:
Boeing 747 Autopilot Pitch Modes Logic (Path Integrator)
I am hoping this will be my final question(s) concerning the 747 autopilot, given that the roll modes are now complete and I am finally on to the pitch modes...
All information on how this autopilot functions is from the following document:
https://ntrs.nasa.gov/api/citations/19730001300/downloads/19730001300.pdf
Okay, so: The Path Integrator.
Integral gains are shown on page 567, and the block diagram showing the integrator implementation is on page 571.
I am struggling to understand just how the path integrator works. Based on the block diagram, I am led to believe that the integrator takes whatever output the pitch autopilot has calculated up to that point, multiplies it by the appropriate integral gain based on mode, integrates, then feeds this into the summing point with the originally calculated output. As an example:
The vertical speed mode proportioner gain is 0.09 degrees pitch command/ft per second of error. Assuming for example a 3 ft/sec error in vertical speed, the commanded pitch change would be 0.27 degrees. Is 0.27 the initial value that the integrator would begin working off of or am I misinterpreting the diagrams? The integral units shown on page 567 are simply listed as "integral/displacement" which seems to me it is integrating based on whatever displacement command the autopilot is currently giving.
Thanks!

Typically the integrator attempts to incrementally adjust the steady-state output so that when the error is zero, the resulting output will be what is required to hold the error at zero.
Think of it this way:
If trying to hold level-flight by controlling the elevator, a non-zero error signal is generated when the flight-path angle is non-zero.  This will cause the elevator to deflect proportinally in an attempt to bring the FPA back to zero- the bigger the error, the bigger the correction.  But! as soon as it is brought back to zero, the error becomes zero again and the elevator input goes back to what it was before. And it was that original elevator position that allowed the FPA to diverge from zero to start with.  This is why an integrator is added. With the integrator, while the error is non-zero, a small percentage of that error is accumulated (integrated) and added to the output- this way when the error goes back to zero, the accumulated error term is still there to get added to the output.  By tuning the percentage added, you can get the accumulated value to come out to just exactly what is needed as a correction to hold your desired FPA.


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What does the term "Course Cut Limit" mean in relation to an autopilot?
In several manuals detailing the autoflight systems of some aircraft, it is common to see the VOR/LOC capture mode have something called a 'course cut limit.' What is this term referring to? I am writing the code for a 747 autopilot based on the gain values given by NASA, and the VOR mode description includes the line "A 34 degree course cut limit is provided."
I have tried googling for this answer but all I could find was some vague definitions saying that this is a guidance control parameter limiting the intercept angle of a capture mode.
So in my example, does this mean that the autopilot will turn my 747 to intercept at no more than a 34 degree angle? If so, at what point does this turn occur? Is it when the capture submode engages? It can't be before that, as before that the autopilot is still referenced to the heading bug while waiting for an intercept/capture.
Thanks!
Chris

Is this a sim autopilot?

Yes, this is a sim autopilot.

After some digging, I came across a swell piece of information in a Qantas 747 maintenance training course about the autopilot, describing the INS capture function:
"When crosstrack distance is greater than 1.5nm, the crosstrack error signal is given a simulated error of 1.5nm until actual crosstrack distance is less than 1.5nm. Combined with a track angle error signal, this limit gives a 25° intercept angle until crosstrack distance is less than 1.5nm, at which point fine capture begins."
In essence, the crosstrack error is limited to 1.5nm, or approximately 9000 feet. The error on the autopilot for crosstrack error in the INS mode is 8.85° bank angle command per 1000 ft of crosstrack distance error. Track angle error is summed into this command at a gain of 3.16° bank per degree of track error. Combined together, this gives an approximately 25° intercept angle when CTD is at the limit of 9000 ft/1.5nm.
After realizing this, everything else clicked.
In the VOR mode, where course cut limit is 34 degrees, beam error is limited to about 4.63 degrees (though capture can begin at a larger beam error of up to 5.8 degrees) so that, combined with course error, the intercept angle is approximately 34 degrees until actual beam error is less than 4.63, at which time full capturing begins.
Likewise, in the LOC mode where the course cut limit is 25°, the beam error is limited to 1.67° (again, capture and intercept happens sooner), so that intercept happens at a 25° angle until actual beam error is less than 1.67°.
Implementation in the sim like this works perfectly. Thanks for your help and input everyone!

This document holds more information on the topic, the document states on page 4:

A course cut limit L was applied to the cross-track bank-angle
input when the airplane was a long distance from the horizontal path

Furthermore a diagram is given on page 16 figure 6, which nicely visualizes this:

The functioning is described as:

When the airplane has a cross-track error
greater than $3\cdot r_t$, the airplane flies toward the path so that $\phi_I = 90^\circ$ The
angle $\phi_I$ decreases linearly from $90^\circ$ to $30^\circ$ as the cross-track error decreases
from $3\cdot r_t$ to $1.5\cdot r_t$.

Thus the "Course Cut Limit" simply limits the authority of the crosstrack limit to the bank angle.
I programmed something similar some while ago, and actually came up with an remarkably similar solution, however only 40 years later :D.

Thank you! I feel as though I am very close to understanding. So in the case of my 747 autopilot, the VOR capture logic works off of beam error and course error. The course cut limit of 34 degrees is applied to the beam error according to my block diagram. I am confused as to how the beam error and course cut limit work together in this instance. Capture begins once beam error is less than 5.8 * sin (course error).

@ChristopherHelton note that the $e_{ct}$ is the crosstrack error, this is the perpendicular distance to the desired path. The course cut limiter is however dynamically scheduled as seen in the document I linked. I am a bit confused however why $\phi_I$ is held constant at $30^\circ$. I do not see how an assymptotic capture is made possible with this limiter...

It seems that the course cut limiter on my 747 autopilot is not dynamically scheduled in any way. I understand that beam error in degrees is essentially equivalent to a crosstrack distance, but no further specification on how the course cut limit works is provided in my documents. I tried an implementation where the beam error signal is not applied until course error is less than the course cut limit of 34 degrees, but this just caused the autopilot to make a strong turn to capture the course, before turning back the other way into an asymptotic capture.

The beam error is not equivalent to crosstrack distance, but rather to the derivative of the $e_{ct}$. Hence the gains $K_y$ (for the crosstrack) and $K_\dot{y}$ for the beam error. This structure represents a PD controller for the crosstrack distance! This is also why you get such nice assymptotical capturing properties. I would just limit the upper output of $e_{ct}$ and forget about the lower limit. Perhaps you can disable the beam error path when the upper limit is active. That should work.

I'm not a software programmer but have flown functional test flights on newly installed autopilot systems.  If you replace the word "cut" with "capture" that makes sense to me. With the Collins systems I operated, I would tune the localizer, engage Heading mode and set a heading using the heading bug to intercept the localizer course, arm the Approach mode, and as the localizer guidance indicator came alive ("off the peg") the Approach mode would activate (capture) the localizer and the Heading mode would disengage. The difference between my heading and the localizer course was limited prior to capture otherwise the aircraft (traveling up to 250 Knots) would pass right through the localizer centerline and require too great a turn radius to successfully execute the approach (bank angles were limited with the autopilot engaged and further dampened or "smoothed" during localizer approach capture mode). a difference of 34 degrees sounds about right for the maximum.

source: Stack Exchange

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Why does directional stability decrease at supersonic speeds?
As an aircraft accelerates through the transonic region from full subsonic to to full supersonic flow its directional or yaw stability decreases. What causes this reduction?

It is a combination of several effects:

Aeroelasticity: With the higher forces at high speed, the structure deforms such that the effective flow angle at the tail surface is reduced.
The supersonic lift curve slope of the tail surface decreases with Mach while that of the fuselage stays roughly constant, so with higher Mach the tail contribution to stability shrinks while the destabilizing contribution of the fuselage stays the same.
Tail location: At supersonic speed the density of air on the upper fuselage is reduced while that on the bottom is increased. Since the vertical sits on top of the fuselage, it flies in rarefied air. Ventral fins, on the other hand, become more effective.

The supersonic lift curve slope is approximately $$c_{L\alpha}=\frac{4}{\sqrt{Ma^2-1}}\cdot\left(1-\frac{\lambda}{2\cdot AR\cdot\sqrt{Ma^2-1}}\right)$$
Note that the second term is for the influence of the triangular surface on the tip where the pressure difference is reduced. Nevertheless, it shows that lift curve slope of the vertical decreases with increasing Mach number.
For the fuselage, the lift curve slope is that of a slender body:$$c_{L\alpha}=\pi\cdot\frac{AR}{2}$$
Note that this equation is independent of Mach, so the destabilizing fuselage contribution stays constant over Mach.
For the beneficial influence of ventral fins see the diagram below which has been published by Lockheed in The F-104 Strafighter Test Pilot's Notebook by Glenn L. Reaves (source):

Thank you. That seems to clear it up nicely.

while the destabilizing contribution of the fuselage stays the same why stays the same? Doesn't the center of pressure move forward?

@Abdullah: The fuselage behaves like a slender body, its center of pressure is already near the tip and the forces on it don't change that much with Mach number.

Are you saying sub vs supersonic or supersonic vs more supersonic? I mean, at subsonic speeds, we have viscosity effects that just don't work that much at Mach>1

I guess the sensitivity to aeroelasticity must vary greatly depending on the relative locations of engine/s and tail fin/s. For example if the tail were rigidly mounted above a single engine, the thrust line would tend to move with it.

@GuyInchbald: Much of supersonic thrust is created in the intake. Yes, fuselage bending will also bend nozzle thrust, but on the fin you need to add the effect of torsion. Using thin supersonic airfoils will give it relatively little torsional stiffness. In total, the destabilizing effect of nozzle thrust adds to the loss of directional stability suffered by the elastic tail, but not much.

Yes indeed. It will still be sensitive to mechanical layout, though.

"At supersonic speed the density of air on the upper fuselage is reduced while that on the bottom is increased." Why is that?

@JZYL Because of compressibility and the desire to fly at an angle of attack in order to create enough lift. Air on top is thinner than air on the bottom of the wings and fuselage. And this becomes worse with higher Mach numbers. Remember that at subsonic speed it's the pressure that changes most while at supersonic speed the density changes become dominant.

Because:

Here you see a fuselage viewed from above. The horizontal line at the right end of the ellipse is the tailfin. On top is a subsonic plane, and at the bottom is a supersonic one.
The thin lines are pressure, the thick ones are forces, and the arrows show flow direction and magnitude.
As you can see, at subsonic speeds suction is more dominant, therefore, the tail, where more suction occurs, produces more sideforce, stabilising the plane.
EDIT: However, this effect is not that significant.
The below diagram shows the tailcone of the same plane.

As you can see, the same thing happens to the tailfin. But the effect of this center-of-pressure movement can't be that significant. However, the back half of the stabilizer becomes less effective at supersonic speed due to being in the low pressure "wake" of whatever is ahead, (the fuselage, as Peter said, or even the front section of the fin itself) as wakes are more defined at these high speeds.
The X-15 had a wedge-shaped tail fin because of this

So if side pressure increases so much at the nose, why does it not also increase in similar proportion at the tail fin?

Let me add another diagram.

@GuyInchbald why does it not also increase in similar proportion at the tail fin? You're missing one point. The destabilizing effect of pressure on the nosecone increases while the stabilizing effect of pressure on the tailcone decreases.

Of course I am missing something, that is why I am asking! Thank you anyway, but I find Peter's analysis to be the more convincing.

@GuyInchbald That's OK, even I upvoted him. I was just clarifying my answer because you misinterpreted it.

Did I? I rather think you must have misinterpreted me to conclude that I was interpreting anything. But you and I have been here before, we should let it pass.

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Is a Mode S transponder required as part of TCAS I?
Is a Mode S transponder required as part of TCAS I?

I assume you are asking if the aircraft with TCAS I installed must also have a Mode S transponder installed as well since that is a requirement of TCAS II?


text:
What information does Mode S provide?
What information about own aircraft does TCAS Mode s( all call) send in the interrogation?
What information does Mode S transponder(intruder)replay to TCAS interrogation?

Are you looking for more information than the wikipedia article can provide?

Closely related

Mode-S provides altitude and identification data, with Automatic Dependent Surveillance Broadcast (ADS-B) adding global navigation data typically obtained from a Global Positioning System (GPS) receiver. The position and identification data supplied by Mode S/ADS-B broadcasts are available to pilots and air traffic controllers.

If my ADS-B OUT in off,   then Mode S provides just altitude and identification (icao number)?

Yes. That's what it says in the answer.


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Which language is used to develop new aircrafts' avionic systems AI?
As an airframe and powerplant maintenance student, I also want to get an avionics license. My main goal is working on artificial intelligence of aircrafts avionics. However I'm concerned about the right language which I should work on. I've been learning Python with OOP for a long time (it was someones advice) but I think I should switch to C++ or Ada.
Honestly, I want to continue with Python but nowadays I have almost no motivation about it. Any advice for my roadmap? Thanks a lot!

Welcome to aviation.SE! Can you tell us exactly what you mean by “AI” in avionics? Is it something that you’ve seen in existing avionics? Or is it something you want to research for future avionics? Or something else? An example would be very helpful. And this question might be relevant.

I hate to burst any bubbles, but power plant maintenance and avionics tech is a very different education/career path than the engineers who design and development avionics.

Hi again. Firstly, after my B1 licenses, I'll take the B2 modules as well. That is not too hard, imo. I want to work on "recognizing objects" and some revelant defense industry. These are existing, as far as I know. They are still in progress, but when I greduate, it wont be a problem, right?

Even after getting the "B2" that is still an aircraft maintenance engineer. If you want to design the actual software that runs it, you will need to attend a 4 year university and major in electrical or software engineering, ideally from a college like Embry-Riddle where you can get a minor in another field like aerospace engineering.

Closest that I find is Google's self navigating car. The code for this car is written in C++. There is no need to use exactly Python for AI, many AI frameworks support multiple programming languages and are actually implemented more or less in C.

Why do you think AI is used in avionics? Or ever would be? Have you looked at what it takes to get safety-critical software certified?

@StephenS: The auto-pilot function definitely is AI. Not the fancy neural-network style Machine Learning (ML) AI that's so much in the news right now, but the classical expert systems AI. However, as Ryudeki specfically refers to " recognizing objects", that's part of the Pattern Recognition (PR) application of AI. All effective PR now uses ML, not expert systems. Getting ML certified for life-critical applications is impossible right now; even getting it to work adjacent to such applications is a challenge.

Aviation is a very verbose world. Pilots do checklists. They use standard phraseology. Each part in an aircraft has a written history like an ancient sword. There are two communicating pilots in the airliner.
And the main selling point of Python has always been its extremely low level of verbosity. Python ommitts lots of declarative code other langages require. This makes the programming fast and easy but does not give context to see the errors early. An error that would prevent Eiffel or C++ code from even being built would often pass ok for Python, and the program will crash during the run instead, and maybe not always, and maybe after it passes the testing.
As a result, I am convinced that Python is not a choice for something that costs lives when it fails. Such cases have they own priorities and "working prototype in two days" is not between them. I think the role of Python in serious aviation will be rather limited.

When developing an AI system (like e.g. a Neural Network), most of the work is done in the training phase. This could still be done in Python, even when implementing the final network would have to be done in Ada or something similar.

It is highly unlikely that a garbage collected language like python would ever be able to be used for something requiring a realtime level of performance assurance. And many systems on an aircraft will need exactly that. I once saw a restrictive standard for C for use in realtime on automobiles, and that would probably be the likely thing here, too.

@Bianfable: Python is far too slow to train Neural Networks. Realistically that's done in CUDA, with Python just acting as a front end. Inference (using the network) can be done in pretty much any language, and if done well is hard real-time.


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How can I control the flight path angle (or the ROC) with throttle in Simulink?
I need to implement an autopilot for my 3DoF (longitudinal motion) aircraft model in Simulink. My model is nonlinear, but linearizing it in Simulink hasn’t been an issue.
I’m trying to develop a control logic that makes my model track a specific flight path angle $\gamma$ using the throttle $\delta_T$ while adjusting speed $V_{\infty}$ using pitch $\theta$ (elevator deflection, $\delta_e$) along the trajectory. I’m struggling to understand how to connect the flight path angle with the throttle. Should I perhaps use this relationship
$$
ROC = V_{\infty}\sin\gamma = \frac{TV_{\infty}-DV_{\infty}}{W}
$$
and linearize it, starting from a trim condition?
Thank you.

Sorry, but writing programs in Simulink/Matlab or other languages is not on-topic on the Aviation forum, no matter what the subject of such programs.

Can I still get some physics advices? Just to find the relationship between those quantities

What you want is a TECS (Total Energy Control System), which controls pitch and throttle simultaneously. This is a classical control scheme for fixed wing aircraft for 3 DoF, see this Google search (unfortunately I have no time to write an answer...)

You can find the classical TECS paper here

@U_flow thank you sir, if you have time you can also write your reply, I can wait. Until then I will read the paper

Instead of using two separate loops for controlling the elevator and thrust, I would recommend to use the TECS (Total Energy Control System) algorithm which simultaneously controls the elevator and the throttle of the aircraft to manipulate the energy state of the aircraft.
The idea behind this algorithm is to control the energy state of the aircraft. One has to realize that the elevator only changes the energy distribution between the kinetic energy (speed) and potential energy (height), while only the throttle introduces or deletes energy! With this in mind, one can easily calculate the energy required for a given airspeed and climb command, calculate the energy distribution between potential and kinetic energy, and then control the elevator and throttle to realize the demanded energy state and distribution.
The energy of an aircraft can be easily described by
$E = m \cdot g \cdot h + \frac{1}{2} \cdot m \cdot v^2$
Dividing by the weight of the aircraft $m\cdot g$ and differentiating once, yields the energy rate of the aircraft per weight:
$\dot{E} =  \dot{h} + V\cdot \dot{V}/ g$
And further dividing by $V$ yields the specific Energy rate
$\dot{E}_s =  \gamma + \dot{V}/ g$
Note that $\dot{h}/V\approx\gamma$ if the small angle approximation is made.
The classical TECS algorithm is displayed below.
The TECS algorithm calculates an specific Energy error (how much energy is lacking) $\dot{E}_{s,c} - \dot{E}_s$ with $\dot{E}_{s,c}$ indicating the commanded specific Energy rate to achieve the commanded descent/climb rate and target acceleration ($\dot{V}$). To do this, a classical PI Controller is used (these are the KTI/s and KTP blocks in the diagram). The result is then given to the throttle command to change the throttle such that the desired energy rate is achieved.
Similarly for the elevator the energy distribution term $L = \gamma - \frac{\dot{V}}{g}$ is formed. Analogously to the throttle command, the pitch command is formed by first calculating a commanded energy distribution $L_c$, forming the energy distribution error $L_e = L_c - L$ and using a PI controller to achieve the desired value.

It should be noted that another control loop for controlling pitch attitude and dampening short period motion should be introduced in the "airplane dependent design" side of the algorithm
This established algorithm was first developed by Lambregts and has since been applied quite successfully in a number of applications. For example by NASA on their Boeing 737 test aircraft or the popular UAV open source control software ArduPilot and Pixhawk PX4.

Thank you for your time sir, I have some questions you can answer whenever you want. The first concerns the situation where I have both a flight path and an acceleration maneuver. According to Lambregts' paper, both the energy rate error and the energy rate distribution error need to go to zero simultaneously. This is achieved via KTP = KEP = 1.0 and KTI = KEI. The latter situation is achieved by matching pitch and thrust responses, but I don't really understand how to do this.

Besides that, I also need to convert the thurst command to throttle command. According to the article, I need a "separate feed forward predictive term" but I've never heard of it. So if you can suggest me where I can learn more, I will really appreciate it.

@leo95nf Concerning your first question. Matching pitch and thrust response means that the control loops managing thrust and pitch, both have a similar time response. It can be a bit different, but that means that it is undesirable that for example the thrust setpoint is reached in 10ms, and the pitch setpoint in 5 seconds.
My advise is to start implementing an example and then itteratively test your controller. It should be relatively straightforward.

Thank you, now is completely clear. The problem is that in my model I do not have any thrust dynamics. I simpy have a thurst model which, given the throttle, the Mach and the altitude, it gives me the thrust. This is implemented as a look up table, so I think I can't continue with this technique

ok maybe I figured out how to do it. I have a pre-def turbofan engine model. It takes in input the throttle and it gives me as output the thrust. So what can I do is to take the thust, birng it in feedback and comparing it the commanded one. This will generate an error that I can feed to a PI controller for example. The dynamics of this closed loop system must be the same one of the pitch attitude, right? So I have just to tune properlly the gains.

@leo95nf Simply insert a first order system with a time response matched to your pitch dynamics behind the commanded throttle to mimic thrust dynamics.

thank you, I appreciate your help. I will let you know!

@leo95nf glad I could help. Please accept this answer if you are satisfied

Hi, sorry if I resume this topic, but I have some issues in tuning the TECS core loop. Do you have any suggestions in determining the gains? Thank you

@leo95nf sorry without knowing the exact details or looking at your model, I think that this is as far as I can help you. Perhaps you have a supervisor or other student you can ask?

can I still get some physics advice?

With aircraft it is far easier to control pitch to the horizon (therefor the aircraft flight path descending, level, or climbing) with throttle control.
The elevator or stabiliator trim is used to set a given airspeed.  This is why we are taught "pitch controls speed, power controls altitude".
In order for this all to work, the aircraft must be staticly stable.  It must have the property of pitching up when it goes faster than its trim speed, and pitching down if it goes slower than it's trim speed.
Once this aerodynamic requirement is satisfied, it is only a matter of power setting to determine angle to the horizon.  Using throttle to do this allows for "fine tuning" of flight path because engine RPM can be adjusted to a high degree of accuracy, instead of struggling to adjust drag with only 3 flap settings.  Power on landings are therefor much easier$^1$.
What should become apparent is that gravity will play a role in aircraft power requirement depending on angle to the horizon, when airspeed and trim setting are constant.
Angle to the horizon is determined by power setting.  If power is insufficient (or none) for level fight, flight path drops below the horizon and gravity assists by the formula W × sin (angle below the horizon).
In a climb, it is opposite, now gravity resists forward airspeed by the same formula.
The amount of Net Thrust to fly the airplane at a given airspeed is always "Gravity thrust" (positive, 0, or negative depending on angle to the horizon) + Propulsive thrust (0 to maximum available thrust).
Gliding at a given airspeed with no power provides the quantity of Net Thrust required for linear steady state flight as W × sin (Glide Angle).
$^1$ except, with a single engine, if the engine is lost you don't make the runway

Commercial solutions to this problem are often pretty complicated but the core is similar to what you're suggesting with thrust target = sin(flight path angle target)*W - predicted drag.
Some catches:

Some ability to account for modelling errors like incorrect weight or drag is essential for speed control.  For example the integral part of a PI controller provides this functionality
since engines can have a slow response time, you may want to have an outer loop for thrust or EPR targets vs fpa, and an inner loop for thrust vs throttle
flight path acceleration can be added to the equation so you don't overestimate the extra thrust needed while you're already speeding up

thank you very much for you answer, can you suggest me any book/paper/video which try to explain this?


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Why did water come into USAirways Flight 1549 if an airplane is sealed and pressurized?
Airplanes are supposed to be air-tight and pressurized.
However, when USAirways Flight 1549 landed on the Hudson river; water started to come into the airplane.
How is this possible if the airplane is sealed? Was there a crack in the fuselage?

I once flew in a Metroliner where the hissing from leaks in the seal of the aft door was louder than the engines. Pressurized fuselages are like sieves. Also, many seals only work one-sided: When internal pressure is higher they are tight, for external pressure they open (think of Space Shuttle SRB seals).

this answer specify they didn't have time to perform all checklist, which may contain items about closing some openings. One comment of this answer specify that the ditch switch was never activated.

Pressurization goes away when both engines were lost. Having the engines submerged certainly doesn't help matters any.

Two reasons:
Airplanes are not completely, perfectly hermetically sealed. They are pressurized using positive air pressure from the engine bleed air and packs. If there is a failure in the system, the aircraft internal pressure will return to the outside ambient pressure. Air will escape the aircraft or enter the aircraft if the outside ambient pressure is greater. Water will react the same way. Especially when you have the combined greater pressure of the water and the atmosphere above it.
Combine this with the structural damage caused by the dynamic pressure of the aircraft impacting the water, the aircraft will have ways of water entering the aircraft through burst seams. The structural and hydrodynamic load alone, created by the rapid deceleration of the aircraft will be enough to tear apart the airframe, open sealed doors and bulkheads, and blow out windows.

Interesting, I always thought these things were sealed like Microwavable vegetables in plastic bags, Thanks for the explanation.

even i didn't see it like that. hmmm

Also, at least some planes use an inflatable seal that would deflate when bleed air pressure is lost.

@Asynchronous - That would make things rather hot and sweaty . Even at high altitudes, the concentration of oxygen is fairly constant, even when the density of molecules is low. We need access to the outside air to facilitate oxygen-carbon dioxide exchange. Lower density just means there are fewer oxygen molecules present to exchange. Also, a certain amount of pressure must be present to allow the lungs to diffuse the oxygen into the blood regardless of how much oxygen is present.

@DeanF.Fascinating analyses. Now I feel smarter. :)

USAir 1549 flooded because the doors opened.  Passengers attempted to open the rear doors, worsening the problem.
Airplanes are not sealed at all.  Not remotely close.  They're not air-tight.  Pressurization is maintained by varying the air outlet, called the outflow valve. A large volume of air is constantly being pushed into the cabin via the engine bleed air valves and air conditioning packs.  A large volume of air is constantly being released from the cabin, meaning the cabin air is completely changed every few minutes.
The aircraft doors use seals which are inflated or held sealed by air pressure.  The metal door doesn't actually seal; flexible seals around the door do that, and those seals work based on air pressure; when the aircraft touches down, it is no longer pressurized; most aircraft must be depressurized before landing, or depressurized to a minimum differential value, typically no more than .5 psid.
In the case of USAir 1549, insufficient time remained to complete the ditching checklist, which included shutting the ditching valve, or avionics exhaust valve, which is done prior to ditching to prevent water from entering.
Flotation of the airframe is largely due to air in the tanks, rather than air in the fuselage.  The airplane isn't designed to float like a boat, and in many cases, a ditching results in a breakup on contact with the water, as water enters pack inlets and other openings and hydraulically tears the airplane apart like a hollowpoint bullet.  The Some aircraft have managed to float for an extended period, but the hope is that it will remain intact and afloat long enough to get out and get away.

There was a hole in the fuselage, and cargo doors had come open. And someone had opened a rear door that couldn't be closed.


text:
Is it possible for squawk to change automatically in emergency situations?
It might be a silly question, but is it possible for an airliner to automatically switch the squawk code to emergency (7700) during a flight? (For example in case of losing all engines)
Thank You In Advance :)

are you asking if it is theoretically possible (i.e. if a system could be designed to do it) or practically possible (i.e. if a system to do it already exists)?

I want to know in situations like ukrainian airlines flight 752 (which was hit by two missiles 3 months ago), if the crew were hit and were unable to control the aircraft, could the squawk change by itself or not?

I'm sorry, but I don't see how that clarifies the point. "could" in which sense? theoretical or practical?

I meant practical , cause I have some data that shows squawk code has changed to 7700 after the aircraft was hit by the first missile , I wanna know whether the crew were alive or not

Whether or not an airlines is in an emergency situation is not a black or white question. There are millions of different internal and external factors that come into play, any combination of which may or may not be considered an emergency. No computer can be programmed to accurately and consistently calculate whether or not a flight is having an emergency situation - in fact, that's one of the main reasons we still have real, live pilots seated up front.
Could you program a computer to detect a few obvious emergencies (total loss of power, rapid decompression, fire etc.)? Sure. But it wouldn't be able to detect everything.
What's more, setting squawk 7700 is not always the best cause of action in an emergency. In fact, most of the time a flight will already be in contact with ATC and have an assigned discreet transponder code - in such a situation, setting squawk 7700 will often have little or no effect, since you can talk directly to ATC on the radio and explain what's going on. You might even risk ATC losing identification if you start changing your squawk code out of the blue, which could add additional risk. In other cases, squawking 7700 can be beneficial, for example if you are not talking to ATC - or maybe the frequency is very busy and you can't get through - or maybe you don't have time to call ATC, but still need their attention (rapid decompression). ATC might also request that you squawk 7700, to reduce workload on their end.
So to summarize, no, a computer would not be able to necessarily detect when a flight is having an emergency situation. And even if it could, setting squawk 7700 might not even be useful.

I am not aware of any commerical transponder that accepts a signal to automatially switch to 7700.  Some aircraft have an "Emergency" switch, that, when set, will do a lot of stuff automatically, like tune the radio to 121.9, activate the ELT, etc.  I'm not aware that any commercial transponder that will automatically switch to 7700.
However, military transponders do have this capability.  When the "emergency" discrete line is "set", the transponder will automatically switch the Mode 3/A code to 7700. One common scenario is to connect this line to the ejection seat system.  When the pilot activates the ejection sequence, the transponder immediately changes the Mode 3/A code to 7700 for "the remainder of the flight," which may not be very long. The logical expectation is the pilot will be in need of rescue and hopefully, the position of the aircraft will register with any ATC entity monitoring the airspace.
The fewer things an ejecting pilot has to do, the better, therefore in this case, the transponder switching to 7700 is automatic in this scenario.
T


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Where do I find charts like this for Europe or Global
I am playing Xplane11 and I am trying to find charts like this
But for Europe or USA and shows every ILS approach, detalied maps etc. For europe I have found one site such as http://www.uvairlines.com/admin/resources/charts-europe.htm but it does not show every airport. Is there a site that shows for any airport ?
Note: I am looking for free sites, not sites like navigraph.

I’m voting to close this question because resource location questions are off-topic as defined in the help center. Also, if you already know about paid services like Navigraph, why would these even exist if there was a free service out there?

@Bianfable the commercial services exist because people will pay for convenience, not because the info can't be obtained for free.

@Bianfable Due to currency I cannot pay that kind of money montly/annualy. Otherwise why wouldn't I use it.

https://skyvector.com/     https://www.airnav.com/

They don't have to be current for simulation and video gaming.  That ought to open up options.

The search term you're looking for is "instrument approach procedure" where sometimes "procedure" will be replaced with "chart" or "plate". Jeppesen provides these commercially for the world in a convenient standardized format for a hefty fee, but the aviation authority for each country also publishes their own versions as part of the ICAO-mandated Aeronautical Information Publication (AIP). In most cases these can be found for free online though it may take some digging.
For the US, the third-party (free) website airnav.com provides a convenient index for the FAA information - search for an airport from their homepage and scroll down for the links under "Instrument Procedures".
Picking a European country at random, Austria's charts are here. Generally you'll want to search for the country's AIP first, then dig through that to look for the instrument procedure charts
I flew myself from California to the UK and back (via Canada, Greenland, Iceland & the Faero Islands) without paying anything for charts this way - paid a lot for fuel though!


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Why do we check Idle operation with carb heat on during a Cessna 172 run-up?
I understand this is to “check our engine is still operating at idle”, but can someone explain this more in depth?
Like why wouldn’t it?

Out of curiosity, which model C172 is this? I've never heard of a carb check at idle before now and I didn't find it in the POHs I have. The only carb check I can see is done during the run-up checks at 1700RPM. But there are lots of different models and it might be specific to a certain engine or whatever.

Both p and m. That is the last check in the run up.

Because, in engaging the carburetor heat, you are decreasing the density of the air entering the engine with the same fuel flow, thereby richening the mixture.  At idle power, if the throttle idle set screw is not properly adjusted, there is a risk of an engine failure from too rich a mixture entering the engine.  The idle power carb heat check is a good way to verify the engine won’t do this in flight.

It also verifies that the carb heat is working. If you pull the lever and the RPM doesn't drop, something is wrong.

That's done at run up rpm. The part you check at idle is the rpm rise from slow idle when taking CH off.

@RonBeyer you would typically verify carburetor heat operation during the run up.  This these here is just an addition to ensure that using carb heat at idle power won’t cause the engine to stall.

Carlo Felicione has given a great explanation of why carb heat at idle would need to be checked. The incoming air heated by the exhaust manifold would be less dense than the incoming air without carb heat. If the fuel mixture is too rich, it may cause the engine to quit. Some may call it flooding the engine.
The most common time for this to be an issue is on final approach. In a Cessna 172, you set the mixture to rich (or best power) to prepare for a possible go around. You would set carb heat to hot in order to avoid carb ice during low power settings. This would be a bad time for engine failure. You are very low to the ground and at a very low airspeed. If the prop stops, you may not have enough airspeed over the prop to continue to windmill it for a restart. And, you may have very little time or mental bandwidth to kick over the starter.
While I have not had this happen on final in real life, I have seen videos of Cessna props stopping when the throttle was brought back to idle in flight. I have also had a 172N prop stop after landing at a field elevation of about 500 feet with carb heat on and mixture full forward. The same N model had no problems for the A&P check. But, the prop stopped later that same week during the run-up at about 500 feet of field elevation.
It performed perfectly during the higher run-up RPM carb heat check. It had a consistent drop in RPMs when carb heat was applied with a consistent return to RPMs when the carb heat was turned off. Idle setting was checked and functioned well at about 700 or so RPMs. The engine quit as soon as carb heat was applied at 700 RPMs. Obviously, that was the end of our flying day. And, I have never skipped the carb heat idle check after that.

Can also second your story. Our aircraft (Not a Cessna) did not have the idle check as part of the checks. What you described happened during a training flight, and the check was introduced by the flying school after


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Why do electronics have to be off before engine startup/shut down on a Cessna 172?
Can someone explain in simple words why we make sure electronics are off before we start up/ shut down our Cessna172 engine?

You could install an avionics master switch to avoid this.

You shut the components off to protect the sensitive electronics from electrical surges that may occur during start up of the engine.
Circuit breakers are designed to trip open if too much current in amps is drawn from the power source by the electrical components or a short circuit downstream of the circuit breaker. They do not necessarily protect the components from too much amperage and/or voltage generated and pushed through the circuit breaker from upstream of the circuit breaker.
To be safe, disconnecting the components from the power source via a physical switch that can be opened Is prudent to mitigate the risk.

The wording here is a bit confusing.  "Upstream" means "towards the source" - I think you mean downstream (ie: circulating reactive currents on the load side network, etc).  In any case, breakers and protectors only trip on sustained overcurrent conditions and the risk during ignition is reactive/inductive voltage spikes generated in the starter that can be large enough to damage electronics without necessarily going into overcurrent conditions, or for long enough for breakers to trip.

The electronic components may or may not draw enough current to trip the breaker at a higher voltage, while still potentially being destroyed. This is the exact same concept as using a surge protector for devices even when a house has circuit breakers.

The main problem lies in the behavior of the coils inside the starter. As all inductors, they build up a magnetic field - this stored energy will cause a high spike in voltage while disconnecting. Though it can be dissipated, there is still a slight chance of damaging sensitive electronic devices. And in aeronautics one does not take chances...

+1 for mentioning the phenomenon better known as inductive kickback, which if not properly controlled can generate high voltages.  These days wiring of large inductors such as motors that are connected to power supplies (including batteries, alternators, magnetos, etc) common to other equipment have "snubber" diodes that are supposed to harmlessly short the voltage spike to ground.  The problem is that snubber diodes sometimes burn out or otherwise fail.

From Edsel Murphy's Laws: "A transistor protected by a fast-acting fuse will protect the fuse by blowing first". One would then suppose that a fuse attached to a whole bunch of transistors, resistors, and etc would be the safest electrical component in the world. :-)

Not familiar with aviation electronics, but can't you just use a reverse-biased diode to get protect from voltage spikes when you have an inductive load?

authors: Bob Jarvis - Слава Україні


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Cessna 208B Caravan various emergencies
looking for help from some experienced Caravan pilots.
So for the 208B Caravan we have our emergency checklist,
which for engine failure tells us to power idle prop feather and fuel lever to idle and basically do a forced landing,
For flameout tells us to either turn ignition on if Ng>50 or put the fuel lever at shutoff and try an airstart,
For FCU malfunction tells us to go power idle and use emergency power lever.
But, how do I know which one of those caused my engine to start quitting?
Don’t all those just result in the torque falling to idle and engine basically quitting?
How do I know which checklist to run?
Thanks for the answers.

I don't have a lot of Caravan time, but I had that same question in training and the way it was explained made sense.  It's best to consider engine failures, roll-backs, and flameouts, as well as your response, in practical scenario-based contexts to ensure the best diagnosis of the cause based on your flight regime as well as the indications.
For example:
SCENARIO 1:  You just flew through heavy precipitation, or maybe hit turbulence at high AOA that interrupted airflow and caused a compressor stall.  If your engine subsequently rolls back, you ought to be thinking flameout first, and should try to catch it quick before Ng decays below 50%.  So, Power to idle to slow fuel flow, and get the ignition switch on right away!  (Give yourself a shot at a quick relight without the inflight checklist.)  If you miss that and the engine seems to be running all the way down, secure the fuel condition lever some time below 50% Ng, and if you have time after setting your glide speed and looking for a landing site, be prepared to try an airstart.
SCENARIO 2:  If the engine just rolls back to idle for no apparent reason you ought to suspect the FCU and engage the EPL (after power to idle...) to regain control.  It may be that you misdiagnosed #1, so in reality one procedure could blend into another.
SCENARIO 3:  This is the mechanical engine failure.  You should really only go straight to this one if it is fairly obvious that it isn't # 1 or 2, because something a bit more dramatic happened.  (I.e. there's no good reason to suspect a mere flameout due to heavy rain or maneuvering, and RPM didn't stop at idle with other conditions indicating normal.)  Something bad has probably happened - maybe there was a loud bang indicating a hard internal failure of some sort, vibration, oil streaking from the cowling, or you just sucked a 25 lb goose down the intake.  You don't want to risk a fire or further damage by leaving it running or attempting a relight if something major has broken, so set your speed to best glide and just shut it down (IDLE, FEATHER, CUT-OFF) then look for a place to land.

Good answer, thanks!

Weren't you driving Caravans right after you retired from the military or something?  Did you ever try to actually run the engine off the emer lever as a demo or anything? I've never heard of that on any other PT-6 powered a/c but the Caravan and Pilatus.

@JohnK, I actually took about 13 years off flying, then just did it part time for a few years until Covid hit. I’d like to get back to it.  I've never used the actual EPL - In the Navy we did a few practice approaches using it in the T-34 simulator. Too much risk of toasting an engine to practice in the airplane, it was pretty touchy as you mention.

I once met the test pilot on the world's second jet airliner, the Avro Canada Jetliner https://en.wikipedia.org/wiki/Avro_Canada_C102_Jetliner, a guy named Don Rogers, when I worked at DeHavilland Canada. The Jetliner was powered with 4 Rolls Royce Derwents, which had typical early fuel controllers (RR had cancelled the engine program for the original engine, and Avro was forced to go with 4 off the shelf Derwents instead of 2 of the newer one - this is 1948).  Like managing a team of very finicky and flighty horses, he said.

With a flameout, the engine stops completely, having flamed out. Torque goes to zero, temps fall off toward ambient.
FCU failures in the PT-6 result in the engine keeping running, but the torque rolls back toward a flight idle baseline, because one of the pneumatic sensing circuits in the FCU malfunctioned and the regulating circuits go off-line, with only the idle circuit in operation.
For the Caravan and PC-12, being a single engine airplane, they added the emergency power lever to mitigate the risk of a single FCU failure putting you into the ground. It turns the engine into a 1940s type of turbine engine, where the fuel controller was very crude, and fuel flow was controlled directly by the pilot (the emer lever is basically a garden hose faucet - your eyes and brain become the fuel control unit).
You now become like a WW2 ME-262 pilot managing his delicate and finicky Jumo turbojets, having to very carefully observe temperatures and torque as you make adjustments, and having to be very gentle with power changes to avoid overtemping or flaming out. But at least you are still flying.

Thank you John. Ok, so Ill identify FCU failure by my power dropping to idle. How would I distinguish between flameout and engine failure for example? Because for flameout Id like to try to restart it using the airstart checklist. For failure, we wouldn’t want to try to restart a failed engine?

I'm not a Caravan driver but I know the PT-6. As I said, it'll still be running, just at idle. If it flames out, torque and ITT will go to zero, oil press lights will come on, all that stuff.  For FCU failure, you implement the emer lever and run the engine with that. The Caravan has an EMER LEVER caution light. I'm not sure if that comes on to indicate FCU fail, or to indicate status when you;ve selected it to run the engine with.  Don't you have a Caravan flight manual?


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Cessna 172S model starting procedure
I just recently started flying a fuel injected cessna after flying carb heated M and P models, and have a question about the starting procedure- particularly about the “warm start”.

when we start the engine for a “warm start”- POH tells us to skip priming with the AUX fuel pump, and have the mixture ICO until the engine starts, at which point we advance the mixture.

My question is- where does the fuel come for the initial start? Shutting down on the previous flight we shut down with cutting the mixture, so there should be no fuel left. And now we are starting with the mixture in ICO position and not priming, so where are the cylinders getting the initial fuel to fire?
Thanks for your answers.

Small nitpick, and I assume you know this: it's not "carb heated" vs. fuel injected, it's carbureted. The carb heat function is tangential to the fuel supply function served by a carburetor or by fuel injectors.

I'm pretty sure what is happening is when you select idle cutoff to shut down, the drop in fuel pressure closes a diaphragm valve in the fuel distribution spider on top of the engine and the immediate fuel flow stops, shutting down the engine.
However, over time, some of the fuel still in the individual lines between the spider and each cylinder drains out the injectors by gravity, the path being down hill. So a small priming charge of fuel, seeping out the injection lines going to the cylinders, is present in the intake ports for some time after shutdown.
Eventually it all finishes seeping and evaporates away, and you're back to a cold engine. But while the engine's still warm, there's just enough fuel there to provide sufficient prime to get the thing to run for a second or two, giving enough time to get the normal system running by pushing the mixture in.
It's a very inexact process, because you don't know just how good a residual prime there is, and when it's time to do additional priming, which if too early will probably flood the engine.
Besides the usual problem of fuel vapour locking in the distribution lines from post shutdown heat, it's one of the things that makes injected engines a challenge to start in hot weather.

just curious where the throttle setting is while all this is going on.  Sort of reminds me of tuning the air mixture screw and the idle screw over and over again on a carburetor.

Unless you think it's flooded, throttle cracked. WOT is for flooded engines and with just a slight prime charge will be too lean and it won't start. Generally, when in doubt, you flood it on purpose (not too much), and crank with WOT waiting for the air/fuel mixture to get to the sweet spot to light off.

Throttle is 1/4” advanced so just very slightly in.


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What is that inertia that makes the nose rise in a spin?
The Flight Training manual mentions
For Ex. 13- spin, that the nose drops due to the loss of lift, and rises due to inertia forces.
Can someone explain to me that inertia force making the nose rise, like Im a 5 year old?

The pitch-up moment comes from the lengthwise mass distribution: Since the airplane swings around a vertical axis, this movement creates an inertial force. For an observer who rotates with the aircraft, this is the centrifugal force of the masses, like in the picture below:

Mass forces in a spin (own work). $\omega_z$ is the rotational velocity around the spin axis, r is the radius between the spin axis and the local mass and m is the mass itself.
The rear parts have a longer radius and, therefore, cause a larger horizontal force at the same rotation speed than the more forward masses. Only if all masses have the same radius (almost possible in an unswept flying wing, or in a vertical dive) will the uprighting moment be zero.

You could say that it's a tilted gyro, spinning about an axis somewhere near or inboard of the main landing gear of the stalled side, and the tendency to spin flat is basically the gyro's erection force is stronger than the aerodynamic forces trying to keep it tilted.

@JohnK It is the usual tendency of a spinning body to change its axis of rotation to one perpendicular to its main inertial axes, as in inertial coupling. Which is the vertical axis for airplanes.

I own a Pazmany PL-2. The Taiwanese built 50 or so of them as a cadet trainer. They discovered they spun flat after about 6 turns. They started a spin test program, doing 300 plus spins on an a/c with a spin chute, trying every bandaid they could think of, strakes, dorsal fins, etc. They thought maybe the tip tanks were doing it, but removing them didn't help. Nothing worked and they gave up.

@JohnK For flat spins the shape of the fuselage nose is the most influential part. It should be flat (see F-5 shark nose). Grob made similar experiences with their G-110 which had a very stable flat spin. The G-115 had a changed fin to cure it from flat spinning.


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Airspeed indication in a forward slip C172?
I have read several topics on the manner with two completely different answers, and just want to make sure I get it right.
When slipping to the right in a C172 (right aileron, left rudder), the static port, located on the left of the fuselage, is slightly blocked from the wind, reducing the pressure in the diaphragm, making the airspeed to OVER-read, correct?

Suggestion: test it and find out and report back.  Stabilize in constant pitch attitude, and report on effects of right versus left rudder inputs, using ailerons as needed to keep wings level.

In a slip, the pitot tube will not be directed into the relative wind, and so the pitot pressure will be lower. This will tend to cause the airspeed to read slower than your actual airspeed.
A decrease in static pressure will cause the airspeed read higher than your actual airspeed. Which of these effects will dominate is not something that can be reliably determined from first principles.
The only thing you can really trust in a slip is not to trust your airspeed indicator!

You did not address the OP's specific question.  MIght there not be a case where the direction of sideslip causes to airspeed indiicator to over-read?  Due perhaps to impingement of the airflow upon the static port?

@quietflyer A reduction of the static pressure would cause the airspeed indicator to over-read, as I say in the first sentence of my second paragraph. However, it is accompanied by the tendency to under-read caused by the reduction of pitot pressure.  Since these two effects compete with one another it's not clear which wins without actually testing it in a plane (or a wind tunnel/simulation, I suppose).


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How to calculate the efficiency of a rubber powered propeller
I made a simple test rig together with a stopwatch to measure the thrust (in grams) with respect to time of a rubber powered propeller. This enabled me to compare the performance of different propellers in terms of duration and average thrust. However, I cannot determine how efficient these propellers are. What could be the missing data that I should take into account to calculate its efficiency?

I think first you have to define efficiency to determine what you're measuring. The canonical definition is $\eta = \frac{P_{out}}{P_{in}}$, and since your prop isn't moving forward on the thrust stand, there is no work being done and thus $P_{out} = 0$ and $\eta = 0$. The upshot is that if you are measuring thrust for the purposes of choosing the best propeller for flight, then you have to test the prop across the airspeed range. If you're only testing static thrust, then your definition of efficiency and your test conditions are wildly in disagreement.

Since an actual model airplane wouldn’t be standing still while flying, I’m wondering if the data you’re taking is relevant.

What a great start to your work!  As others suggested, measure torque to wind the prop (plotting resistance to winding per turn).  First see how consistent the rubber band is at storing energy (does it fatigue after say 20 windings)).  Next you need to test efficiency at the desired cruising speed of your airframe.  So:  take it to the hill and see which prop makes you plane fly farthest at the speed you want (if you don't have a wind tunnel).  For these types of aircraft, a larger slower prop is generally more efficient, but a faster one may work better due to Reynolds number factors.

You should avoid dated units, such as the gram-force. Now, the area under your graph is in $MLT^{-1}$ units, force x time. These correspond to the variation of momentum $∆MV$ imparted to the air by the prop. Perhaps you could use that area as a reference, when testing other props...

In order to compare efficiency, you would need some way of measuring the torque exerted by the rubber band, and compare (thrust generated/torque required). You could construct a second arm&scale mechanism at 90º of the existing one that is loaded by the wound up rubber band.
Momentum theory has a glossary efficiency term, the Figure of Merit. Basically, for helicopters, it is defined as (Ideal power to hover/actual power required to hover), with the actual power being the power from induced drag + power from profile drag.

Chapter 2.8 of Principles of Helicopter Aerodynamics by J. Gordon Leishman contains a detailed explanation, The figure above is Fig 2.8 of the book, with thrust coefficient $C_T$ defined as
$$C_T = \frac{T}{\rho \cdot A \cdot (\Omega R)^2}$$
With T = thrust[N]; A = disk area [m$^2$]; $\Omega R$ = blade tip speed [m/s].
Fig 2.8 shows the dimensionless thrust coefficient $C_T$, and the chapter mentions that only rotors (or propellers) of equal area should be compared. Or compare $C_T$ with $C_Q$ for different propellers, with $$C_Q = \frac{Q}{\rho \cdot A \cdot \Omega^2 \cdot R^3}$$ with Q = torque in [Nm].
The above should get you going, more detailed info in the reference books.


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Can winglets be used to compensate the high induced drag created by a low aspect ratio wing?
I have read in some articles that the major advantage of having a high aspect ratio (AR) wing is to reduce the induced drag. If winglets main job is also to reduce the induced drag, how much drag does it reduces? Would it be possible for a low AR wing with well-designed winglets to have the same induced drag as a high AR wing?

Related: Is a winglet better than an equal span extension?

In the lens of potential theory, we can explain and predict the efficacy of winglet through how much it dilutes the trailing vortices strength. Just like we can show that the Elliptical lift distribution produces the least induced drag for a given span length on a flat planform, we can equally derive the optimal lift distribution for any wing+winglet combination of arbitrary winglet height and cant angle.
The graph below, collected from Drela, Flight Vehicle Aerodynamics, shows parametrically how the most optimal induced drag of various winglet extensions and cant angles compare against the flat Elliptical drag, if the total lift on the wing is fixed. $h_{winglet}$ is the vertical height of the winglet above the planform plane, and $\theta_{winglet}$ is the cant angle measured from the planform plane (0 means a simple span extension, while 90 means a perpendicular winglet).

Image from Drela, Flight Vehicle Aerodynamics
As you can see, a simple span extension ($\theta_{winglet}=0$) is always better. For example, to achieve the same induced drag reduction as a span extension of 20%, we must have a vertical winglet whose combined height is 40% span (assuming we don't allow span extension whatsoever).
I must stress that the results above are a comparison between the most optimal planform + winglet combinations. Optimality means that when the winglets are attached, the wing itself is modified to achieve optimality. In practice, with the wing itself fixed, you could only do worse.

Consider two wings of equal area, one with half the span and twice the chord so it has a quarter the aspect ratio and four times the tip losses.
Now add winglets, which are sized according to the chord, so the low-span wing will have winglets twice as tall. The winglets recover say 50% of the tip losses. Although the large winglet recovers more efficiency as an absolute value of drag, the short-span wing lost so much in the first place that it still loses twice as much as the high-aspect-ratio wing.
So no, winglets are no substitute for aspect ratio. In typical applications they allow a modest reduction in span, of roughly the height of the winglet on each wing.


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Will it be more efficient if airplanes flap like birds?
Many airplanes have parts that are designed to mimic some features of animals. I have also read that birds are incredibly more efficient than most of the airplanes we see today. I wonder, will the future passenger airplanes resemble a big flapping bird? Considering that in the future we might have a lighter and more durable materials for wings.

3 people voted to close this! Amazing.  Maybe this site should split into an aeronautical engineering section and those who wish to fix everything with software patches.

@RobertDiGiovanni as you can see, the reason for closing is that another question already have answers that can answer this question. If you look at the linked question, answers speak in a concise way of ornithopters, square-cube-law, already attempted flapping wing designs, speed consideration... I think this is a dupe because element in answers are already tackled in the other question.

Yet Aaron's picture is priceless.

In short, the answer is No. For many centuries, many proto-aeronautical engineers believed that the secret to manned flight would be found in complex flying machines that mimicked the flapping techniques of birds. These ornithopter designs were doomed from the start.

Aside from the obvious issues involving strong, rigid, durable, and light materials, there are tremendously complex psycho-muscular factors involved in biological flight mechanics that are impossible to duplicate with machinery. A bird has the benefit of nerve endings all over its body which provide immediate and intimate feedback of every minute shift and change. Additionally, the bird has fine motor control over its flight surfaces to a much greater degree than any aircraft can realistically replicate, and all at the speed of thought.
The good news is, efficiency can be relative. While the deft flight of a bird can be graceful, beautiful, and evolutionarily effective for its avian lifestyle, it will certainly never accomplish a trans-Pacific trip at 500 knots. Mankind has learned a lot from watching our feathered neighbors, but ultimately, our journey among the clouds lies along a different path from theirs.
That said, some small drone applications show promising possibilities for mimicking insect wings, so mother nature may still have a few more tricks to teach us.

I strongly disagree with the second paragraph. If we assume lighter more durable materials, we can also safely assume use of multiple integrated sensors and neural network computing to control the aircraft. No biggie with current thechnology. I do, however, agree that it is not, and will not be more efficient to flap the wings of an aircraft. It is best to separate lift, propulsion and control into separate entities. In future, control may be integrated into wings (no separate control surfaces), but thats still decades away from being commercially viable.

@Jpe61: I’m sorry that you disagree, but every square inch of skin can contain over 1,000 separate nerve endings, each sending entirely unique information which is then processed and responded to in a fraction of a second. The idea of mechanically replicating the complexity of a biological creature to this degree is pure science fiction speculation at best. We can barely get robots to achieve bipedal ambulation on a flat surface due to the complexity involved - a task which many biological creatures handle with ease. Introduce uneven terrain, and suddenly most robots fall flat.

Our brains do not continuously process data from every nerv ending in our body. There are special filters, so to say, that block most of the data from ever reaching even the reflex level of our nervous system. Imagine the processing power needed to fully monitor everything our senses are subjected to; simply impossible. As for bipedal ambulation, or, as we common ppl say, walking, pls check out Boston Dynamics, namely their two-legged (or bipedal, as the more civilized would say) Atlas. Pretty fancy stuff he does for a robot.

@Jpe61: I never used the word brain. I said that the information from the nerves was processed and responded to. There is not even an argument to be made here. I know of Boston Dynamics work, I’ve watched videos of Atlas wipe out on simple obstacles dozens of times, even as his own creators explain the tremendous difficulty of getting a machine to accomplish what animals can easily do. There are no humanoid robots doing parkour. No humanoid robots climbing El Cap. How much more difficult for avian flight? Biology excels at adaptation. Programming doesn’t even come close.

@Jpe61: Snicker at the scientific terminology all you like. Walking is unspecific. Cats walk. Ants and spiders and centipedes walk. “People-style-walking”, or bipedal ambulation, immediately let’s you know what category of locomotion we are talking about. I’m not trying to rain on your sci fi parade - just pointing out that billions of years of development has led to extremely sophisticated methods of movement that far outshine our best attempts to replicate them.

Animal like distributed neural network computer could be utilized in the hypothetical case presented in the question. The reason animals, such as humans, have such great amount of sensor all over and inside their bodies is not because they are needed for controlling movement. Walking, or flight control in the case of birds, would be possibe with far less dense sensory network. Totally doable in near future i say. But again, just not worth it. Oh, and it seems you have not seen the latest videos of Atlas doing his trics.

@Jpe61: It sounds like you've got a pretty good handle on it. Can't wait to see what you come up with.

Don't hold your breath

Likely no but let's see
The relevant research study can be found here. In this study, it was shown that for some insects, the induced power can be twice the ideal value in hover, and can be even higher than this in forward flight. This is worse than possible with the fixed wing.
From the other side, the same study says that flapping  wings do not show classical abrupt stall characteristics  at high angles of attack. This is due to the leading-edge vortex on the top surface of the flapping wing. Thus, the flapping wing  achieves significantly higher lift coefficient at high angles of attack and may be more effective, even if less efficient.
Flapping wings can provide hovering as well when fixed wing cannot.
Also, propeller powered insects do not exist so it is not clear how the control experiment is done. If versus aircraft, the size difference may matter. A 3 mm propeller or jet engine for a mosquito insect may not be a more efficient solution.

Several things come to mind:
Probably not.  A flapping wing is only producing thrust during half its operating cycle (except hummingbirds and insects, who's wings generate thrust in both directions when hovering) so there is energy wasted in the wing's recovery cycle in normal forward flight.  It might be better for the bird to keep its wings "fixed" and use another oscillating surface to provide the thrust, like a fish tail, but nature has apparently decided this is not a viable engineering solution for making horizontal thrust in air and in balance, it's better to oscillate the lifting surface.
Efficiency aside, the big show stopper is that a mechanical flapping wing machine, to the extent it can be perfected, will always suffer from the the problem of the body moving vertically in an inertial response to the wings being driven up and down.  You would not want to ride is such a machine vary long.
The main difference between the natural world and the man-made one is the man-made one's ability to assemble separate parts and make them operate together to convert energy potential to force without touching (either elements separated by an oil film, or a rolling interface) to produce power in various forms (crankshafts, connecting rods, pistons, turbine rotors, etc).  Living beings must produce their thrust from elements that are physically interconnected in a single organism grown from a single egg.  This is a huge engineering limitation for mother nature and is why nature couldn't invent the wheel.
So I would say, as an engineering exercise, ornithopters are actually sub-optimal, and nature and evolution produces ornithopters because that's the only design solution available when the "design toolbox" so to speak is limited to what can be done with a single interconnected organism, and to the extent that a man made version can be replicated, nobody would want to spend more than 2 minutes in the thing anyway.

The tail is evolutionary lost in modern birds, but ancient birds (Archaeopteryx) had the feathered tail. For some reason it did not evolve into the main thruster like for a fish. Probably flapping wing has some advantages.

Yeah i expect that to work with any efficiency the oscillating tail would need to be nearly as large as the wing, defeating the purpose.

Note that the most efficient birds mainly flap to take off, otherwise they use air currents to soar.  The albatross even has a locking mechanism so that its wings stay extended without using any energy: http://blogs.bu.edu/bioaerial2012/2012/11/17/flying-without-flapping/

Get the book called The Miracle of Flight by Stephen Dalton.  He takes you on a journey through the Reynolds number scale from insects to birds to aircraft.  Written for the layperson, with terrific photography.

It is amazing how ingrained into the psyche of many aviation enthusiasts, even today, is the issue of lifting the plane "as birds" when the issue of propulsion against drag is predominant as speed increases.
So, reduction in drag becomes the key issue.  Indeed, a flapping wing is perfect for a relatively slow, lightly wing loaded bird as it "moves a lot of air a little at a time" much more efficiently than a duck holding its wings straight out and paddling franticly with its webbed feet (which would work fine in higher density water).  But a bird can flap its wings due to much lower wing loading, which allows to lift and thrust with the same device, and lower airspeed, which allows for a much lighter structure from relatively lower aerodynamic stresses.  Early biplanes with very large propellers took a similar approach until...
Increasing speed placed emphasis on the need for increased thrust.  The higher wing loading of larger aircraft require higher speed to fly.  So, at the expense of efficiency and weight savings, aircraft trend towards greater but less efficient thrust, partially offset by the ability to fly higher, giving a higher True Air Speed.
But recently, with the incorporation of high by-pass fans (and higher fuel prices), designers are returning to the old wisdom that birds employ so well.


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Which among the two contributes most of the lift, the upper or lower surface of the airfoil?
I see some explanations that lift is due to the suction of low pressure on the top surface of the wing and deflecting down of airflow on the bottom which results to upward force lift. I wonder, which among the two surfaces (top and bottom) contributes most of the lift?

It is not really a "Suction" effect as such, whilst there is a pressure drop, I've seen it explained this way.
Newtons Third Law states that for every action there is an equal and opposite reaction.
This means that as the wing interacts with the airflow, it imparts a force upon that airflow directing the airflow downwards towards the ground. In an equal and opposite direction, the airflow now imparts an upward force upon the wing, lifting the aircraft off the ground.
I think this is an easier way to describe and understand lift.

I've experimented with removing the top or the bottom.  In my experience it is better to keep the top and remove the bottom, so I think the top contributes the most lift.

Lift is just the vector summation of ALL the pressure pushing on the wing. And the air on the top of the wing is pushing it DOWN. We only get lift up because the air on the bottom of the wing is pushing the wing up harder than the air on the top is pushing it down. So much misconception here. Suction !!???? !  What? do you think the little air molecules have vacuum cleaners? or is it some kind of glue?

In subsonic flight, the majority of the lift is generated by the lower pressure (compared to ambient) at the upper surface of the airfoil, which produces a net suction. A portion of the lift is also provided by the lower surface that has higher pressure, but is much smaller in comparison.
The graph below shows experimentally derived data on a symmetric airfoil, which breaks down the pressure distribution between upper and lower surfaces. The vertical axis of the graphs shows difference between local static pressure and free-stream static pressure divided by dynamic pressure, which is also called pressure coefficient. Note that a negative number means lower pressure compared to free-stream static pressure, and is suction; a positive number means higher pressure, and is pushing.
As you can see, in the 9deg AOA case, the upper surface suction dwarfs the lower surface push force by observing that the area under the curve is much larger for the former.

Image ref: http://web.mit.edu/13.021/13021_2003/Lifting%20surfaces/lectureC.htm
The next two plots show the stall and post-stall. This airfoil stalls at leading edge, so the entire suction peak is removed once the flow is stalled.

There is NO SUCH THING as negative pressure,(or suction, for that matter). This is just as silly as saying that there is such a thig as Negative heat when the temperature is is a negative number, or that all the people in Austraila are living in a Negative earth cause their latitude is a negative number. All real pressure is positive. It is caused by moving molecules bouncing off the surface, and imparting their momentum to the object they are bouncing off of. The only reason we use negative numbers to measure it is because we are comparing it to atmospheric pressure (14.7 psi)

Lift is just the vector summation of ALL the pressure pushing on the wing. And the air on the top of the wing is pushing it DOWN. We only get lift up because the air on the bottom of the wing is pushing the wing up harder than the air on the top is pushing it down.

@CharlesBretana Of course there's no negative pressure. But as you said, there's negative pressure differential. Suction is a phenomenological and industry-wide technical term. If the top wing generates most of the pressure differential for the given area, then it is absolutely technically correct to call it suction. Amended my answer to clarify the plots.

But the use of the word suction creates a wrong understanding. The ops question demonstrates this. Nothing is "pulling" on the top surface of the wing. The air on the top is "pushing" down on the wing. Use of the word suction creates the wrong mental image.


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What is the modern approach to extract vehicle system responses to control input?
What is the modern approach to extract vehicle system responses to control input?
Is everything thrown into a CFD simulation and the responses recorded or is something else done? Are wind tunnels still relevant?
Real time CFD computation isn't happening on planes so there has to be some simpler dynamics figured out for the controller to use.

I do not understand the question. What is "extract vehicle system responses to control input"? Who is "extracting" it? To do what? Why do you mention "real time CFD"?

Am I correct in deciphering this as "How are aircraft responses to control inputs determined currently during design phase"? I assume you wondering whether wind tunnel testing is still used or is everything done with computer simulations?

OP: For the non aerospace engineers here, is CFD "computational fluid dynamics"? If so, please edit that into your question.

Why would you need to do fluid dynamics computations in real time? This question doesn't make sense at the moment.

I think I understand your question, but I am not sure. Please provide some context, what is the problem you want to solve?  Do you want to model how an aircraft responds to control inputs, so that you can design a control system/autopilot? Do you want to do this while the aircraft is still on the drawing board? Or at a later stage, when a prototype has been built?

One approach to control is optimize the model in real time(needs to be faster than real time),  This is probably not happening in flight computers for CFD simulations.

Are you asking about how vehicle responses to control system inputs are modeled for the purposes of design, or for the purposes of actually controlling the aircraft?

Both, there has to be a reduced order model to describe response to control inputs, so how is it found?

The goal of system identification is to supply a model for flight dynamics simulation. Such a simulation is very useful for design of flight control and flight guidance systems as well as for performance evaluation amongst others.
The way such a flight simulation is built up typically, is by characterizing the forces and moments acting upon the aircraft with respect to airspeed, height, angle-of-attack etc. but also in respect to flap and control surface deflections.
It is important to note, that especially the change of these parameters in respect to the aforementioned parameters determines how the aircraft behaves dynamically. Therefore for example the exact frequency of the dutch roll or other dynamic effects of the aircraft. You can find more information on this under the search term stability or control derivatives.
Several ways exists to determine these forces and moments:

Wind tunnel tests: This is the most accurate and widespread way to determine these forces and moments without having to build the aircraft (Building the whole aircraft might not only be costly, but in some cases you cannot fly the aircraft without having first designed a suitable flight control computer, because your aircraft might be unstable). A scale model of the aircraft is built, and then tested with the help of a force balance in order to directly measure the forces and moments under various conditions.
Computer fluid dynamics: Instead of physically building the aircraft for real-life testing or wind-tunnel testing, you model the aircraft for testing inside a virtual wind tunnel. Note that you typically do not let the aircraft fly "free" but rather you try to simulate the aircraft under various conditions again in order to determine the forces and moments given the aforementioned parameters in order to find out the stability derivatives. This minimized the computation time. This approach of course is much cheaper then wind tunnel testing, but because computer fluid dynamics is not 100% accurate, there always remains some uncertainty.
In-flight system identification: For older aircraft designs or in order to validate the findings which you determined via the aforementioned system identification methods, you can determine the forces and moments/stability derivatives in flight. Again it is not always possible to fly your vehicle, either because you still want to itterate the design or because you just cannot because the aircraft lacks a flight control computer. However, this is of course the most accurate way to determine the forces and moments, as flight tests are the "real deal" containing all "dirt" and real-life effects without any assumptions.
First principles: For a first evaluation, you can try to roughly determine these derivatives from first principles. While not very accurate it might give you a good enough idea for a first guess/simulation/design itteration.

Once you obtained a set of forces and moments/stability derivates over your parameter space (airspeed, height, angle-of-attack, sideslip angle and control deflections) then it it possible to simulate the aircraft with the help of the equations of motions of an aircraft in order to subsequently either design a flight control system, or calculate performance metrics etc. etc.

Thanks, do you know how movement would be considered? For example as you change a rudder the aircrafts direction will change and perhaps it will no longer fly in the direction it is pointed. I don't know anything about aerodynamics but I'm somewhat competent in multivariate control.

@FourierFlux Sure, you would take the equations of motion (i provided a link), plug in the control derivatives, (linearize them if needed) and simulate the Differential equations as needed.

If you are only interested in lateral movement (involving angle of sideslip and yaw angle) you can subdivide the equations of motion in vertical/height movement as well as lateral movement. Normal books on aircraft system dynamics provide the appropriate concepts


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How do you compute the net lift force location?
There are a lot of formula that give an approximate solution to the lift and drag forces associated with certain wing configurations.
What I'm curious about is how do you find where the net force is applied assuming a rigid body?
EX you're interested in designing a plane, you need to make sure the net lift force from the wings is applied to the COM so the plan stays level with minimal input.

If not otherwise specified, it is given in respect to the MAC

Would that be the center of gravity?  Keep in mind drag will also cause pitching moments, as well as the thrust vector. (Usually down, as adding power can cause the plane (aerodynamicly) to pitch up).


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Lift without aerofoil?
I have seen some RC planes flying without any form of airfoil, the wing is literally a flat board.
What I want to understand is how lift is generated under these conditions. I understand kinetic energy is transferred from air onto the wing, pushing it up(assuming your aircraft is angled up) but I don't understand why air from above the wing doesn't cancel it out.

Not the same question.

Have you ever stuck your flat hand out the window of your car on the freeway and played around with twisting it?

What is the performance of a flat plate wing? Of course this depends on the Reynolds Number (thus on the scale), as explained in the answer.

When we study the lift produced by an airfoil, we add together the lift caused by the camber (the curved mean surface of the airfoil) and the lift caused by the angle of attack.
Any symmetrical airfoil (say a NACA 0012, or your flat board of an airfoil) will have zero camber, so that contribution to lift goes away.
However, the contribution due to angle of attack remains.  It does not care that the shape is just a board.
The reason we prefer a nicely rounded airfoil is to delay or avoid flow separation.  A flat board has sharp corners at the front that can encourage separation -- by increasing the leading edge radius of the shape, we can delay that.

So what causes the lift then? Does there end up being lower air pressure above the wing due to a delay caused by time it takes to fill the void above the wing? I don't understand why upward lift isn't canceled out by downward push

In my experience, explanations of lift can be: simple, correct, intuitive -- but you only get to pick one.  There are many discussions on ASE diving into this fundamental question.


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How do you calculate the optimal cruising speed of an airfoil?
I am working on a school project (aircraft design).
I would like to know how to calculate the cruising speed of a given airfoil. For example, if I have a NACA 5613-35 airfoil (that is the NACA 4 digit modified series), what is the the cruising speed at which this airfoil has the best aerodynamic properties?
Another way to ask this question: If I want my aircraft to have a cruising speed of 120 knots, how do I choose an airfoil to fit the cruising speed requirement?
I tried looking it up on google and on some other aviation forums, but I did not find the answer.
Thanks in advance and have a great day!

You need to go through airfoil tech-sheets. These are available ini some textbooks

Correct version: You need to go through airfoil tech-sheets. These are available in some textbooks - AC Kermode's "Mechanics of Flight" has everything you need. Combination with fuselage and tailplane is also to be understood. Also, in aviation, and theory of flight, "the best aerodynamic properties" can have more than one answer - best range, best endurance, high speed, more payload and these often conflict with each other.

The cruising speed is only one parameter. For optimum flow conditions, you need to look at the Reynolds number and the lift coefficient at which the airfoil shows its best behavior and then tailor the aircraft to achieve those numbers at its design operating speed. Normally, a higher Reynolds number is better, but at low speed (= low Reynolds numbers) airfoil performance drops off a cliff. It is important to stay above this limit in order to get good performance.
This tailoring is done by selecting the wing area in order to achieve the right wing loading (the ratio of airplane mass to wing area) and the wing's aspect ratio (that is the ratio between span and chord) to arrive at the lowest possible drag. Selecting too high an aspect ratio (wing becomes too slender) will drive up the mass of the wing structure and reduce its Reynolds number. Conversely, picking too little wing span will give you plenty of wing chord, but induced drag will be higher than necessary.


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Can Airbus A320 doors be opened mid-flight?
Today I learned how airliner doors are typically an outward facing plug and cabin pressure would make them nearly impossible to open mid-flight.
It's easy enough to see that for the shape of Boeing 737 doors, for example.
But looking at this video of the A320 door operation, it doesn't exactly look like it forms a plug.  So if it were unlatched while the cabin is pressured, would it get pushed open?

related: Are type-C doors plug doors? and What fail safe protections exist to prevent non-plug type cabin doors opening in flight?

It's a plug door.  The door nestles behind an array of cleats and a set of L shaped guide tracks and has to move up some distance, as you can see in the video, to clear the cleats and exit the L track.  When it's dropped into the locked position, it's not dependent on any moving parts to resist pressure because it settles behind the fixed cleats in the outer frame.  The cleats in the door bear on the ones in the frame to resist outward pressure and the door is opened by lifting it to mis-align the cleats so the door cleats clear the frame cleats. That makes it a plug door.
It looks like there are other levers and cams to assist with the door pull-in, and probably some form of locking pin(s) to block vertical movement when latched. If the door was, say, 36 inches wide by 80 inches high, at an 8 psi cabin pressure differential, there is about 23000 lbs pushing on it, so you won't be moving it unless you are The Hulk.
The CRJ Regional Jet main pax doors are an example of non-plug doors that just pull in flush on a simple hinge at the bottom, and latching shear pins extend into holes in the door frame.  This door is dependent on the shear strength of the movable pins, and the shear loads when pressurized pretty much weld the pins in place.

Are you sure those are plug-doors? I thought A320 family doors were type-C which are not plug doors.

It operates the same as the door in the first video, with the same L tracks and perimeter fittings, but a bit smaller with different contours. It has to move up to clear a set of fixed cleats in the frame.  When closed, pressure is pushing the fittings on the door against the fittings of the frame, making it a plug door.  The fact that that skin surface of the door is not larger than the fuse opening is irrelevant. The structural load paths are door statically pushing against fuselage.On the CRJs, the right side service door works the same as these doors, but the main pax door is a non-plug...

...that depends mostly on 6 shear pins operated by a complex assembly of levers and bellcranks, and is a nightmare to rig.  That door just moves straight out with no vertical or lateral movments because the shear pins, a pull-in cam at the top, and the piano hinge at the bottom hold the door against cabin pressure, and the shear pins completely retract when opening to allow the door to swing straight out.

Does that mean the answers in the link question was wrong 5 years ago?

I would say so.

I figured there's something subtle about the way that door works!  This makes we wonder if it's more prone to leaking air pressure than a 737 door.

Depends on the type of seal system being used.  You can often see inflatable seal systems that use bleed air for inflation, that retract into a groove in the door frame and extend out when inflated to bridge the gap.  The RJ's door seal is a passive tubular gasket that presses against a ridge in the frame perimeter, with little vent holes that are open to the cabin side of the ridge, so cabin pressure provides some positive inflation pressure to push the tube against the ridge in the frame.  Works OK and doesn't require an inflation system.  Leaking doors seals can make quite a racket.

@JohnK I'm confused. Your answer begins by stating the A320 is fitted with plug doors while it seems to be fitted with type-C doors which are not plug doors. What am I missing?

I was randomly googling this again and I did find reports of doors violently opening due to residual cabin pressure straight from the horse's mouth: https://safetyfirst.airbus.com/preventing-violent-door-opening-due-to-residual-cabin-pressure/  So I guess it's possible up to a certain point, if not at full cabin pressure.

In that link it talks about checking an inward opening door to make sure there is no cabin pressure.  On a lift-up outward opening plug door, I imagine that when the pressure is down to a fraction of a psi, you can work the linkage of a lift up door with the leverage of the actuating lever, then once it's clear that residual pressure is enough to fling it out.

If the door is not larger than the opening, I don't see anything that prevents the door from being opened when the pressure pushes the door out and the operator doesn't need to pull it in first.

@RonenFestinger there are lug projecting from the frame that align with mating lugs on the door when it drops into place. When the door is raised the lugs are offset so the ones on the door can get past the ones on the frame.

@JohnK but what prevents it from being opened when in air? The pressure prevents the door from being pulled in, not from being pulled up and out.

The door is being pushed against the lugs attached to the fuselage. It has to move up so each lug on the door clears its mate on the fuselage.  But you won't be able to slide the door up off the retaining lugs with the leverage provided by the actuating system when there is 8 psi pushing on 3000 square inches. 11-12 tons or so. And the linkage may require the door to move in slightly, make it even more impossible.  Some doors aren't even plug doors, they're held closed by shear pins. You can't work the mechanism to withdraw the shear pins when the pressure differential is present.